The CDF of multiple order statistics Let $X_{(1)}\geq X_{(2)}\geq\cdots X_{(K)}$ be the order statistics of the random variables $X_1,\,X_2,\,\ldots,X_K$, which are independent and identically distributed exponential random variables with parameter 1. 
Now let $Z_k=\frac{Y_k}{X_{(k)}}$, where $Y_1,\,Y_2,\,\ldots,Y_K$ are also independent and identically distributed exponential random variables with parameter 1.
I wish to find the cumulative distribution function (CDF) of $\max\limits_{k}\{Z_k\}$, which is given by
$$\text{Pr}\left[\max\limits_{k}Z_k\leq z\right]$$
How can I find the above CDF?
The problem is that $\{Z_k\}_{k=1}^K$ are not independent, because the denominator is an order statistics. I tried to condition on the denominator and then average as
$$\begin{split}
&\underbrace{\int\int\cdots\int}_{K-\text{Fold}}\text{Pr}\left[\frac{Y_1}{\alpha_1}\leq z, \frac{Y_2}{\alpha_2}\leq z,\cdots, \frac{Y_K}{\alpha_K}\leq z\right]f_{X_{(1)}, X_{(2)}, \ldots, X_{(K)}}(\alpha_1, \alpha_2, \ldots, \alpha_K)\,d\alpha_1\cdots d\alpha_K\\
=& \underbrace{\int\int\cdots\int}_{K-\text{Fold}} \prod_{k=1}^K \text{Pr}\left[\frac{Y_k}{\alpha_k}\leq z\right]f_{X_{(1)}, X_{(2)}, \ldots, X_{(K)}}(\alpha_1, \alpha_2, \ldots, \alpha_K)\,d\alpha_1\cdots d\alpha_K
\end{split}$$
but got stuck in evaluating it. 
Thanks in advance for any tips.
 A: Define $\tilde{Z}_{k} = \frac{Y_{k}}{X_{k}}$ and $Z_{k} = \frac{Y_{k}}{X_{(k)}}$ with $Y_{k}$ and $X_{k}$ as in the question.
The random vectors $[\tilde{Z}_{1},\tilde{Z}_{2},...,\tilde{Z}_{K}]$ and $[Z_{1},Z_{2},...,Z_{K}]$ will have different distributions, however...
The distribution of the order statistics $\tilde{Z}_{(K)} = \max\limits_{k}\frac{Y_{k}}{X_{k}}$ and $Z_{(K)} = \max\limits_{k}\frac{Y_{k}}{X_{(k)}}$ will be equivalent since the ordering in the denominator is "undone" by introducing new random variables and taking the maximum.
So, to find the density of $Z_{(K)}$, it suffices to find the density of $\tilde{Z}_{(K)}$.
Distribution of $\tilde{Z}_{k}$
The CDF of the ratio of two exponential distributions with rate 1 is $F(z) = (1+z^{-1})^{-1}$, for $z>0$. See relevant question here.
Distribution of $\tilde{Z}_{(K)}$
$\tilde{Z}_{(K)}$ has CDF: $Pr(\tilde{Z}_{(K)} < z) =F(z)^{K}$ by rules for order statistics. So $Pr(\tilde{Z}_{(K)} < z) = (1+z^{-1})^{-K}$.
This is also the CDF of the maximum you are after.
Note: I'm using the standard definition of order statistics where $X_{(i)} \leq X_{(j)}$ for $i < j$.
